Video: Kinematics and Forces

In this video we learn about the kinematic equations and how to use and derive them, as well as the forces that create accelerating objects whose motion is described by these equations.


Video Transcript

In this video, we’re going to learn about kinematics and forces. We’ll learn about objects in motion described through the kinematic equations, and we’ll see how forces help to create and explain this motion.

To start out, imagine that you’re participating in a stunt at an airshow. You’re in a plane flying straight ahead with a speed 𝑣 in line with a bull’s-eye target ahead on the ground. You want to find out what is the horizontal distance along the ground from the center of the bullseye called 𝑑 at which you can release a ball from the plane which will fall and land in the center of the bull’s-eye.

You know the plane is an altitude 𝐴 above ground and that as it falls the ball will encounter a wind which pushes on the ball with the force equal to half the ball’s weight. Given all this information, to solve for 𝑑, we’ll want to know something about kinematics as well as forces.

Kinematics is a field that describes object motion under constant acceleration. And when we think of object motion, we can consider that to be an object’s acceleration, velocity, and displacement. To describe an object’s motion, we have four equations, all of which are in terms of acceleration, velocity, displacement, and time.

Looking over these four equations, we see that two of them involve solving for 𝑣 sub 𝑓, an object’s final velocity, and two involve solving for 𝑑, an object’s displacement. But the particular algebraic form of each equation doesn’t limit what variable we solve for. For example, considering the equation at the bottom left, we can rearrange this equation to solve for any of the four variables involved, depending on the information we had and the information we wanted to solve for. That’s true for all four of the kinematic equations.

Considering these equations, the question may come up of where they come from. These equations do have an origin. They’re derived from the definitions of the terms involved in the equations, acceleration, velocity, and displacement.

For example, the definition of acceleration 𝑎 is that it’s equal to the change in velocity over the change in time. We can rewrite this expression to say instead of Δ𝑣, that that change is equal to a final velocity minus an initial velocity. Then, if we multiply both sides of the equation by Δ𝑡, the time difference, that expression cancels on the right-hand side. And if we add the initial velocity 𝑣 sub zero to both sides, that term also cancels on the right-hand side. We’re left with an expression that says the final velocity is equal to the initial velocity plus 𝑎 times the change in time.

This equation is the same as the one written in the top left of our four. We won’t derive the other three kinematic equations. But this example- we won’t derive the other three kinematic equations, but all of them do come from these basic definitions of acceleration, velocity, and displacement.

When using the kinematic equations practically, we want to be sure of two things. First, the kinematic equations apply only when acceleration is constant over whatever motion or process we’re considering. Over any interval where acceleration is not constant, that is it changes, these equations are not valid. Second, because of the directionality involved in positive-negative velocity, acceleration, or displacement, we want to be careful to establish those directions at the outset of a problem and carefully follow them as we go through a solution. That’s a little bit about kinematics. Now, let’s talk about how kinematics work together with forces.

Forces connect with the kinematic equations because forces, we know, create motion. In particular, when a force on an object is unbalanced, that object accelerates. It’s Newton’s second law of motion that tells us how an object accelerates under an unbalanced force. This second law is a vector equation, meaning that the direction of the net force is the same as the direction of the object’s acceleration. So long as the net force acting on an object is constant, so is that object’s resulting acceleration, which is then the acceleration that we use in our kinematic equations to solve for object motion.

As they’re written, these equations of motion express magnitude quantities, speeds, acceleration magnitudes, and distances. But we know that, in general, acceleration, velocity, and displacement are vectors. In these equations, the vector nature of the terms involved is preserved and maintained using positive and negative signs. That’s part of why establishing a sign convention in a kinematics and forces example is so important.

Putting this altogether, let’s consider an approach for solving problems involving both forces and kinematics. First, if we’re not given the net force acting in our particular scenario, a free body diagram can help us find that. Once we have that net force, the second law of motion can let us use it to solve for the object’s resulting acceleration 𝑎.

Once we’ve collected all of the motion-based information we can gather, we choose positive and negative directions for those terms, velocity, acceleration, and so forth. Then, looking over the list of kinematic equations, we choose the one or ones that will help us solve for what we want to find. Let’s practice using this approach through an example.

An object is acted on by three simultaneous forces. 𝐅 one equals negative 3.00𝐢 plus 2.00𝐣 newtons, 𝐅 two equals 6.00𝐢 minus 4.00𝐣 newtons, and 𝐅 three equals 2.00𝐢 plus 5.00𝐣 newtons. The object experiences acceleration of 4.23 meters per second squared. The object is initially at rest. Find the mass of the object. Find the speed of the object after 5.00 seconds. Find the components of the velocity of the object after 5.00 seconds.

Given the three forces 𝐅 one, 𝐅 two, and 𝐅 three acting on an object initially at rest, which result in an acceleration of 4.23 meters per second squared, we want to solve first for the mass of the object. We can label that 𝑚. We also want to solve for the speed the object attains after 5.00 seconds. We’ll call this speed 𝑠. And finally, we want to solve for the velocity components of the object after that same amount of time. We’ll call these components 𝑣 sub 𝑥 and 𝑣 sub 𝑦.

To start on our solution, we can draw a sketch of the forces acting on this object. If we sketch in the three forces acting on our object on a pair of axes in units of newtons, we see that we can add these together algebraically to solve for the net force acting on the object. Writing out 𝐹 one, 𝐹 two, and 𝐹 three by their component parts, when we add them together, we find a net resultant force of 5.00𝑖 plus 3.00𝑗 newtons.

Now that we know the net force, we can use Newton’s second law, which says this net force equals the object’s mass times its acceleration to solve for object mass 𝑚. By the second law, object mass is equal to the magnitude of the net force acting on it divided by its acceleration 𝑎. We solve for the magnitude of the net force by squaring each of its two components, adding the results together, and taking the square root of that sum.

Dividing that magnitude by our acceleration, given as 4.23 meters per second squared, we find that object mass, to three significant figures, is 1.38 kilograms. That’s the mass of the object under the influence of these three forces.

Next, we want to solve for the speed of our object after a time of 5.00 seconds has elapsed. Since the acceleration, in this case, is a constant value across that time, this tells us that the kinematic equations of motion apply to this object’s motion. We look through these different equations seeking a match between what we’re trying to solve for, final speed, and the information we already know.

Since we’re told that time, which we can call 𝑡, is 5.00 seconds, the first equation written down is a match for what we want to solve for. Written in terms of our variables, this equation is 𝑠 is equal to the initial speed of the object plus its acceleration times time. We know that because it started at rest, its initial speed is zero. So, its final speed is simply 𝑎 times 𝑡, or 4.23 meters per second squared times 5.00 seconds. This is equal to 21.2 meters per second. That’s the object’s final speed after 5.00 seconds under this acceleration.

Lastly, we want to solve for the components of the velocity of the object at the same time 𝑡 equals 5.00 seconds. To start on that solution, let’s draw the net force acting on our object in on our graph. Our net force, the sum of 𝐹 one, 𝐹 two, and 𝐹 three, is pointed at some angle, we can call it 𝜃, from the positive 𝑥-axis. This angle is important because the direction of our net force by Newton’s second law is the same as the direction of the resulting acceleration. And since our particle started from rest, that means its velocity will also be in the direction of force since it’s in the direction of acceleration.

Here’s what we know so far about our final velocity at 𝑡 equals 5.00 seconds. We know, first of all, that its magnitude is equal to 𝑠, which we’ve solved for. And second, we know that this velocity has an 𝑥- and a 𝑦-component. But we just don’t know what those components are. However, if we solve for the direction of the net force, then we can figure out how the final speed magnitude divides up over the two components of velocity.

So, looking at our diagram, if we draw in the 𝑥- and 𝑦-components of the net force and recognize we know their numerical values from solving for them earlier, we can write that the tangent of our angle 𝜃 is equal to 3.00 newtons divided by 5.00 newtons. Taking the arctangent of both sides, we now have an expression for the angle 𝜃. And when we solve for it, we find it’s roughly equal to 31 degrees.

Now, that we know the angle 𝜃, we can solve for the components of velocity based on the final magnitude of that velocity, which is speed. We can write 𝑣 sub 𝑥 as the magnitude of the velocity, which is 𝑠, times the cos of 𝜃 and 𝑣 sub 𝑦 as 𝑠 times the sin of 𝜃. Plugging in for these values when we solve for 𝑣 sub 𝑥 and 𝑣 sub 𝑦, we find they’re equal, to three significant figures, to 18.1 meters per second and 10.9 meters per second, respectively. In component form then, our final velocity is 18.1𝑖 plus 10.9𝑗 meters per second.

Let’s take a minute to review what we’ve learnt about kinematics and forces. We’ve seen here that force imbalances are what accelerate objects and that acceleration relationship is found through Newton’s second law. We’ve also seen that when an object’s acceleration is constant, its motion can then be described by what are called the kinematic equations. These four equations are based on the relationships between acceleration, velocity, and displacement.

We’ve also gotten practice using a problem-solving strategy for examples with kinematics and forces. First, we’ve solved for the net force involved in a scenario. Then, we apply Newton’s second law of motion to get information about its acceleration. Then, we established which directions are in the positive and which directions are in the negative. And finally, we apply the kinematic equation or equations that let us solve for the variable we’re seeking. Knowing this process and these four equations are helpful for understanding kinematics and forces.

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